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Homework Help
Engineering and Comp Sci Homework Help
Determine the Reactions at O and the Cable Tensions
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[QUOTE="Northbysouth, post: 4103532, member: 430702"] [h2]Homework Statement [/h2] The light right-angle boom which supports the 520-kg cylinder is supported by three cables and a ball-and-socket joint at O attached to the vertical x-y surface. Determine the reactions at O and the cable tensions. [h2]Homework Equations[/h2] [h2]The Attempt at a Solution[/h2] I'm a little unsure if what I should do with this question. When I spoke with my TA, he said to find the unit vectors of the cables, multiply those unit vectors by their appropriate tensions (which are unknown) and then to take the cross product of these with the vector from O to a point of action on the line. 0 = r[SUB]OA[/SUB]XT[SUB]AC[/SUB][itex]\hat{}n[/itex][SUB]AC[/SUB] + r[SUB]OB[/SUB]XT[SUB]BD[/SUB][itex]\hat{}n[/itex][SUB]BD[/SUB] + r[SUB]OB[/SUB]XT[SUB]BE[/SUB][itex]\hat{}n[/itex][SUB]BE[/SUB] - 520*9.8 My calculations are as follows: [itex]\hat{}n[/itex][SUB]AC[/SUB] = <-1.25, 1, -2.60>/[itex]\sqrt{}9.3225[/itex] [itex]\hat{}n[/itex][SUB]BD[/SUB] = <0, 2, -2.60>/[itex]\sqrt{}10.76[/itex] [itex]\hat{}n[/itex][SUB]BE[/SUB] = <0, 0, -2.60>/[itex]\sqrt{}2.60[/itex] T[SUB]AC[/SUB][itex]\hat{}n[/itex][SUB]AC[/SUB] = T[SUB]AC[/SUB]<-1.25/[itex]\sqrt{}9.3225[/itex], 1/[itex]\sqrt{}9.3225[/itex], -2.60/[itex]\sqrt{}9.3225[/itex]> T[SUB]BD[/SUB][itex]\hat{}n[/itex][SUB]BD[/SUB] = T[SUB]BD[/SUB]<0, 2[itex]\sqrt{}10.76[/itex], -2.60/[itex]\sqrt{}10.76[/itex]> T[SUB]BE[/SUB][itex]\hat{}n[/itex][SUB]BE[/SUB] = T[SUB]BE[/SUB]<0, 0, -2.60/2.60> r[SUB]OA[/SUB]XT[SUB]AC[/SUB][itex]\hat{}n[/itex][SUB]AC[/SUB] = <0, 0, 2.60> X <-1.25T[SUB]AC[/SUB]/[itex]\sqrt{}9.3225[/itex], T[SUB]AC[/SUB][itex]\sqrt{}9.3225[/itex], -2.60T[SUB]AC[/SUB]/[itex]\sqrt{}9.3225[/itex] r[SUB]OA[/SUB]XT[SUB]AC[/SUB][itex]\hat{}n[/itex][SUB]AC[/SUB] = <-2.60T[SUB]AC[/SUB]/[itex]\sqrt{}9.3225[/itex], -3.25T[SUB]AC[/SUB]/[itex]\sqrt{}9.3225[/itex], 0> r[SUB]OB[/SUB]XT[SUB]BD[/SUB][itex]\hat{}n[/itex][SUB]BD[/SUB] = <1.80, 0, 2.60> X<0, 2T[SUB]BD/[itex]\sqrt{}10.76[/itex][/SUB], -2.60T[SUB]BD[/SUB]/[itex]\sqrt{}10.76[/itex]> = <-5.2T[SUB]BD[/SUB]/[itex]\sqrt{}10.76[/itex], 4.68T[SUB]BD[/SUB]/[itex]\sqrt{}10.76[/itex], 3.60T[SUB]BD[/SUB]/[itex]\sqrt{}10.76[/itex]> r[SUB]OB[/SUB]XT[SUB]BE[/SUB][itex]\hat{}n[/itex][SUB]BE[/SUB] = <1.80, 0, 2.60> X <0, 0, -T[SUB]BE[/SUB]> = <0, 1.80T[SUB]BE[/SUB], 0> R[SUB]OF[/SUB] = <0.90, 0, 2.60> R[SUB]OF[/SUB] X <0, -5096, 0> = < 4586.4, 0, 0> M[SUB]Ox[/SUB] = -2.60[SUB]TA[/SUB]/[itex]\sqrt{}9.3225[/itex] - 5.2T[SUB]BD[/SUB]/[itex]\sqrt{}10.76[/itex] + 4586.4 M[SUB]Oy[/SUB] = -3.25T[SUB]AC[/SUB]/[itex]\sqrt{}9.3225[/itex] + 4.68T[SUB]BD[/SUB]/[itex]\sqrt{}10.76[/itex] + 1.80T[SUB]BE[/SUB] M[SUB]Oz[/SUB] = 3.60T[SUB]BD[/SUB]/[itex]\sqrt{}10.76[/itex] Thus I get: T[SUB]AC[/SUB] = 5385.98 T[SUB]BD[/SUB] = 0 T[SUB]BE[/SUB] = 3185 But this doesn't make any sense. T[SUB]BD[/SUB] can't be zero. Help would be greatly appreciated. Thanks [/QUOTE]
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Determine the Reactions at O and the Cable Tensions
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